4
$\begingroup$

I have been learning about derived categories from Hartshorne's "Residues and Duality". One of the main theorems, as it is presented there seems to be the canonical isomorphism $RF \circ RG \cong R(F \circ G)$ that happens under various circumstances.

Questions:

1.) Suppose I have an additive functor $F$ between abelian categories (say with enough injectives and projectives). If $F$ is left or right exact I can consider its right or left derived functors respectively. With the derived category formalism is there a useful notion of a derived functors if $F$ is neither left nor right exact? Perhaps such a notion could associate to a short exact sequence a long exact sequence (that continues in both directions).

2.) I've heard that with the formalism of derived categories one is able to compose left and right derived functors. I assume that you have to work in an unbounded derived category somehow, because right and left derived functors are defined on different derived categories. This seems to be entirely omitted from Hartshorne. Is it useful to consider such compositions (say in algebraic geometry, or etale cohomology)?

References will be greatly appreciated!

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ You can consider the right and left derived functors of an arbitrary additive functor, not necessarily left or right exact; IIRC it will just in general fail to be the case that the zeroth derived functor coincides with the original functor. $\endgroup$
    – Qiaochu Yuan
    Commented Apr 9, 2013 at 5:56
  • $\begingroup$ An example of a "useful" composition of left and right derived functors is the direct image of D-modules, which is a $R\pi_*$ applied to a derived tensor product. $\endgroup$
    – Ketil Tveiten
    Commented Apr 9, 2013 at 8:30

1 Answer 1

Reset to default
1
$\begingroup$

For 1) you should look into Positselski papers. For 2) see this question Composing left and right derived functors.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Sasha, so it seems like #1 isn't mainstream. Is that right? $\endgroup$
    – LMN
    Commented Apr 9, 2013 at 6:23
  • $\begingroup$ It depends on what do you want from the derived functor. If you want it to be a natural extension to the derived category of the initial functor, you have to be careful. As far as I understand in case of a nonexact functor the answer is that you have to modify the derived category appropriately. I think you can find an explanation in Positselki papers. $\endgroup$
    – Sasha
    Commented Apr 9, 2013 at 6:41
  • 3
    $\begingroup$ Specifically, double-sided derived functors are discussed in my book "Homological algebra of semimodules and semicontramodules: Semi-infinite homological algebra of associative algebraic structures" (the arXiv version can be found at arxiv.org/abs/0708.3398 , but the printed book version is better). A more precise reference is Section 2.7. $\endgroup$
    – Leonid Positselski
    Commented Apr 9, 2013 at 10:32
  • 3
    $\begingroup$ This is a definition of a balanced double-sided derived functor of a functor of two arguments, in which the objects/complexes whose substitution at the first argument makes the functor exact in the second argument are used as resolutions. The construction is sensitive to the fine details of the definitions of the derived categories from which the would-be derived functor is supposed to act, and a modification of the conventional definitions is sometimes required, as Sasha mentions above. See Section 0.2.3 for counterexamples. $\endgroup$
    – Leonid Positselski
    Commented Apr 9, 2013 at 10:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .